In figure, AB = AC, D is the point in the interior of ∆ABC such that ∠DBC = ∠DCB. Prove that AD bisects ∠BAC of ∆ABC.

Given: AB = AC, D is the point in the interior of ∆ABC such that ∠DBC = ∠DCB.
To Prove: AD bisects ∠BAC of ∆ABC.
Proof: In ∆DBC,
∵ ∠DBC = ∠DCB ...(1) | Given
∴ DB = DC    ...(2)
| Sides opposite to equal angles of a triangle are equal
In ∆ABC,
∵ AB = AC    | Given
∴ ∠ABC = ∠ACB
∴ ∠ABC = ∠ACB    ...(3)
| Angles opposite to equal sides of a triangle are equal
Subtracting (1) from (3), we get,
∠ABC - ∠DBC = ∠ACB - ∠DCB
⇒ ∠ABD = ∠ACD    ...(4)
In ∆ADB and ∆ADC,
AB = AC    | Given
DB = DC    | Proved in (2)
∠ABD = ∠ACD | Proved in (4)
∴ ∆ADB ≅ ∆ADC
| SAS congruence rule
∴ ∠DAB = ∠DAC    | CPCT
⇒ AD bisects ∠BAC of ∆ABC.

Given: In ∆ABC, AB = AC, ∠A = 36°. The internal bisector of ∠C meets AB at D.
To Prove: AD = DC

Proof: In ∆ABC,
∵ AB = AC
∴ ∠ABC = ∠ACB    ...(1)
| Angles opposite to equal sides of a triangle are equal
Also, ∠BAC + ∠ABC + ∠ACB = 180°
| Angle sum property of a triangle
⇒ 360° + ∠ABC + ∠ACB = 180°
⇒    ∠ABC + ∠ACB = 144° ...(2)
From (1) and (2),

Now, ∵ CD bisects ∠ACB

Again, In ∆ACD,
∠ACD = ∠CAD (= 36°)
∴ AD = DC
| Sides opposite to equal angles of a triangle are equal

Given: AB = BC, AD = EC
To Prove: ∆ABE ≅ ∆CBD
Proof: In ∆ABC,
∵ AB = BC    | Given
∴ ∠BAC = ∠BCA    ...(1)
| Angles opposite to equal sides of a triangle are equal
AD = EC    | Given
⇒ AD + DE = EC + DE
⇒ AE = CD    ...(2)
Now, in ∆ABE and ∆CBD,
AE = CD    | From (2)
AB = CB    | Given
∠BAE = ∠BCD    | From (1)
∴ ∆ABE ≅ ∆CBD | SAS congruence rule.

Given: X and Y are two points on equal sides AB and AC of a ∆ABC such that AX = AY.
To Prove: XC = YB
Proof: In ∆ABC,
∵ AB = AC    | Given
∴ ∠ABC = ∠ACB    ...(1)
| Angles opposite to equal sides of a triangle are equal
Again, AB = AC    | Given
AX = AY    | Given
Subtracting, we get,
AB - AX = AC - AY
⇒    BX = CY    ...(2)
In ∆BXC and ∆CYB,
BX = CY    | From (2)
BC = CB    | Common
∠XBC = ∠YCB    | From (1)
∴ ∆BXC ≅ ∆CYB
| SAS congruence rule
∴ XC = YB    | CPCT

Given: A triangle ABC in which AB = AC
To Prove: ∠ABC = ∠ACB



Construction: Draw the bisector AD of A so as to intersect BC at D.
Proof: In ∆ADB and ∆ADC,
AD = AD    | Common
AB = AC    | Given
∠BAD = ∠CAD
| By Construction
∴ ∆ADB ≅ ∆ADC
| SAS congruence rule
∴ ∠ABD = ∠ACD    | CPCT
⇒ ∠ABC = ∠ACB
Yes, the converse is true.